Four charges Q_1, Q_2, Q_3, and Q_4 of the sa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the force exerted by charge $Q_i$ on $q$ be $\vec{F}_i$. The $x$-component of the force from a pair of charges at $\pm x_0$ is proportional to

Vedclass · Accepted Answer

$P-3, Q-1, R-4, S-2$

Vedclass · Answer

(A) Let the force exerted by charge $Q_i$ on $q$ be $\vec{F}_i$. The $x$-component of the force from a pair of charges at $\pm x_0$ is proportional to $(Q_{left} - Q_{right})$. The $y$-component is proportional to $(Q_{left} + Q_{right})$.$(P)$ All positive: $Q_1=Q_2=Q_3=Q_4 = +Q$. The $x$-components cancel out due to symmetry. The $y$-components add up, resulting in a net force along $+y$ (Option $3$).$(Q)$ $Q_1, Q_2$ positive, $Q_3, Q_4$ negative: The $y$-components cancel out. The $x$-components add up in the $+x$ direction because the positive charges are on the left and negative on the right, pushing $q$ to the right (Option $1$).$(R)$ $Q_1, Q_4$ positive, $Q_2, Q_3$ negative: The $x$-components cancel out. The $y$-components are negative because the charges closer to the origin $(Q_2, Q_3)$ are negative and stronger in their $y$-effect than the farther positive charges $(Q_1, Q_4)$, resulting in a net force along $-y$ (Option $4$).$(S)$ $Q_1, Q_3$ positive, $Q_2, Q_4$ negative: The $y$-components cancel out. The $x$-components result in a net force along $-x$ because the net effect of the charges on the left is negative and on the right is positive, pulling $q$ to the left (Option $2$).Thus, the correct matching is $P-3, Q-1, R-4, S-2$.

List-$I$	List-$II$
$P. Q_1, Q_2, Q_3, Q_4$ all positive	$1. +x$
$Q. Q_1, Q_2$ positive; $Q_3, Q_4$ negative	$2. -x$
$R. Q_1, Q_4$ positive; $Q_2, Q_3$ negative	$3. +y$
$S. Q_1, Q_3$ positive; $Q_2, Q_4$ negative	$4. -y$

Similar Questions

Three identical charges are placed at the three corners of an equilateral triangle as shown in the figure. Which of the following statements is true for the electric field $E$ and electric potential $V$ at the center $O$?

The dimension of $\frac{1}{2} \varepsilon_0 E^2$,where $\varepsilon_0$ is the permittivity of free space and $E$ is the electric field,is:

Two equal negative charges $-q$ are placed at points $(0, a)$ and $(0, -a)$ on the $Y$-axis. A positive charge $q$ is released from rest at point $(2a, 0)$. What will be the motion of this charge?

Vedclass Test Series

Exam Paper Generator

Online Exam Module